ECE-656 Fall20171

NAME:______________________________________ PUID:______________________________________

ECE656Exam3SOLUTIONS:Fall2017

December12,2017MarkLundstromPurdueUniversity

Thisisaclosedbookexam.Youmayuseacalculatorandtheformulasheetattheendofthisexam.OnlytheTexasInstrumentsTI-30XIISscientificcalculatorisallowed.Therearefourequallyweightedquestions.Toreceivefullcredit,youmustshowyourwork(scratchpaperisattached).Theexamisdesignedtobetakenin75minutes.BesuretofillinyournameandPurduestudentIDatthetopofthepage.DONOTopentheexamuntiltoldtodoso,andstopworkingimmediatelywhentimeiscalled.Thelasttwopages,whichlistequationsmayberemoved,ifyouwish.100pointspossible,25perquestion

1)25points 2a)5points 3a)5points 4a)10points

2.5ptseach 2b)5points 3b)10points 4b)5points 2c)10points 3c)5points 4c)10points

2d)5points 3d)5points

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ECE-656 Fall20172

Exam3Solutions:ECE656Fall20171a) Whenwewritetherecombinationterminthevariousbalanceequationsas

Rφ = nφ − nφ

0( ) τφ ,sometimesatermcorrespondingto nφ appearsandaterm

correspondingtoitsequilibriumvalue, nφ

0 ,doesnotappear.Whendoesitnotappear?

a) Understeady-stateconditions.b) Underspatiallyuniformconditions.c) Whenthebalanceequationcorrespondstoamomenthigherthan2.d)Whenthebalanceequationcorrespondstoamomenthigherthan3.e)Whenthequantityinthebalanceequationisaflux.

1b) Whenwewriteadrift-diffusionequationintheform,

Jnj = nqµnE j + 2 3( )µn ∂W ∂x j ,whatassumptionarewemaking?

a) Non-degeneratecarrierstatistics.b) Thetemperaturedoesnotvarywithposition.c) Theelectrontemperatureisequaltothelatticetemperature.d) Thekineticenergyisequallydistributedbetweenthethreedegreesof

freedom.e) OnlythattheBTEisvalid.

1c) Whatdoesmomentequationdoes

φ !p( ) =υx p2 2m*( ) giveus?

a) Thecarriercontinuityequation.b) Thecarrierfluxequation.c) Thecarrierenergybalanceequation.d) Thecarrierenergyfluxequation.e) Thecarrierenergysquaredcontinuityequation.

1d) Whensimulating

r t( ), p t( )( ) ,inphasespacebytheMonteCarloapproach,whichofthefollowingistrue?

a) r t( )

iscontinuousand

p t( ) iscontinuous.b) r t( )

isdiscontinuousand

p t( ) iscontinuous.c) r t( )

iscontinuousand

p t( ) isdiscontinuous.d) r t( )

isdiscontinuousand

p t( ) isdiscontinuous.e) Noneoftheabove

ECE-656 Fall20173

Exam3Solutions:ECE656Fall20171e) Whatis“selfscattering”inaMonteCarlosimulation?

a) Amanybodyeffectinwhichanelectroninteractswithitself.b) Anelectron-electronscatteringeventinwhichanelectronscattersfromanother

electron.c) Anelectron-electronscatteringeventinwhichanelectronscattersfromthe

entireplasmaofalltheelectrons.d) Amathematicaltechniquethatsimplifiesthecomputationoffree-flight

times.e) Amathematicaltechniquethatsimplifiesthecomputationofthefinalscattering

state.

1f) Inpractice,onecommonlyextendsthenear-equilibriumdrift-diffusionequationto

high-fieldsbyreplacingthemobilityanddiffusioncoefficientsbyelectricfielddependentquantities,asin Jnx = nqµn E( )E x + qDn E( )dn dx .WhatassumptionisnecessarytowritetheDDequationinthisform?

a) Parabolicenergybands.b) Non-degeneratecarrierstatistics.c) Themicroscopicrelaxationtimeapproximation.d) Thattheenergyrelaxationtimeisshorterthanthemomentumrelaxationtime.e) Thattheshapeofthedistributionattheparticularlocation,whateveritis,

dependsonlyontheelectricfieldatthatlocation.1g) Intheclassicdescriptionofthevelocityvs.electricfieldcharacteristicinbulkSi,

υd =µn0E 1+ E E C( )2

,approximatelywhatisthemagnitudeof E C ?

a) ≈ 0.1kV cm .b) ≈1kV cm .c) ≈10 kV cm .d) ≈100 kV cm .e) ≈1000 kV cm .

ECE-656 Fall20174

Exam3Solutions:ECE656Fall20171h) Whatismeantbytheterm,“non-local”semiclassicaltransport.

a) TransportthatcannotbedescribedbyaDDequationwithafield-dependent

mobilityanddiffusioncoefficient.b) Steady-statetransportinanelectricfieldthatvariesmorerapidlyinspacethan

theenergyrelaxationlength,where Te istheelectrontemperature.c) Transienttransportinanelectricfieldthatvariesmorerapidlyintimethanthe

energyrelaxationtime.d) Alloftheabove.e) Noneoftheabove.

1i) Underwhatconditionsdoesvelocityovershootoccurforarapidlyvaryingelectric

field?a) Whentransportisballistic.b) Whentransportisquasi-ballistic.c) Whenthemomentumrelaxationtimeismuchshorterthantheenergy

relaxationtime.d) Whenthemomentumrelaxationtimeismuchlongerthantheenergyrelaxation

time.e) Whenthemomentumrelaxationtimeisnearlyequaltotheenergyrelaxation

time.1j) Whichofthefollowingistrueaboutaballisticdevicewithtwo,idealLandauer

contactsatdifferentvoltages?

a) ThedistributionfunctioninthedeviceisaFermi-DiracdistributionwiththeaverageFermilevelofthetwocontacts.

b) ThedistributionfunctioninthedeviceisaFermi-DiracdistributionwiththeFermilevelofthecontactwiththemorepositivepotential.

c) ThedistributionfunctioninthedeviceisaFermi-DiracdistributionwiththeFermilevelofthecontactwiththemorenegativepotential.

d) Eachstateinthedeviceisinequilibriumwithoneofthetwocontacts.e) Eachstateinthedeviceisinequilibriumwithboththetwocontacts

ECE-656 Fall20175

Exam3Solutions:ECE656Fall20172) MonteCarlosimulationsofhigh-fieldtransportinbulksiliconwithanelectricfieldof

100,000V/cmshowthefollowingresultsfortheaveragevelocityandkineticenergy:

υd = υ = 1.04×107 cm/s

u = KE = 0.364 eV Forthisproblem,youmayassumeasimpleparabolicbandwithaneffectivemassof

m* = mC

* = 0.26m0 .2a) Estimatetheaveragemomentumrelaxationtime, τ m .Anumericalansweris

required.Solution:

µn = υ E = 104 cm2 /V-s

µn =

q τ m

mc*

,where mc

* istheconductivityeffectivemass.UseMKSunitsforthecalculation:

τ m = q

µnmc* =

µnmc*

q= 104×10−4 × 0.26× 9.11×10−31

1.6×10−19 1.54×10−14

τ m = 1.54×10−14 s

2b) Estimatetheaverageenergyrelaxationtime, τ E .Anumericalansweris

required.Solution:

Fromtheenergybalanceequation:

JxE x = nq υ E x =

n u − u0( )τ E

τ E =

u − u0( )q υ E x

u0

q= 1.5

kBTL

q= 0.039

τ E =

u − u0( )q υ E x

=0.364− 0.039( )

1.04×107 ×105 = 0.313×10−12

τ E = 0.3 ps

ECE-656 Fall20176

Exam3Solutions:ECE656Fall20172c) Estimatethedriftenergyinelectronvolts.Anumericalanswerisrequired.Solution:

Edr =

12

m*υd2 = 0.5 0.26( ) 9.11×10−31( ) 1.04×105( )2

= 1.28×10−21 J

Edr eV( ) = 1.28×10−21

q= 8.01×10−3 eV

Edr eV( ) = 8.01×10−3 eV

2d) Estimatethethermalenergyinelectronvolts.Anumericalanswerisrequired.Solution:

Fromtheformulasheet: u = 1

2m*υd

2 + 32

kBTe

32

kBTe = u − 12

m*υd2 = 0.364− 0.008 = 0.356

32

kBTe = 0.356 eV

Virtuallyalloftheenergyisthermalenergy.

ECE-656 Fall20177

Exam3Solutions:ECE656Fall2017 3) Whenderivingthemomentumbalanceequationin3D,atensor,

Wij =

1Ω

υi p j2

f r , p,t( )p∑

occurs.Thisquestionisaboutthediagonalcomponentofthetensor, Wxx .Youmayassumeasimple,parabolicenergybandandspatialvariationinthex-directiononly.

3a) Usethebalanceequationprescriptiontoobtainanexpressionforthex-directedflux

associatedwith Wxx .Solution:

Fφx ≡

1Ω

12

m*υx2⎛

⎝⎜⎞⎠⎟υx f x, !p,t( )

p∑ = FWxx

.

FWxx

isafluxofthequantity, Wxx .Itcanalsobewrittenas

F

Wxx

≡ 1Ω

12

m*υx2⎛

⎝⎜⎞⎠⎟υx f

p∑ = n

12

m* υx3

3b) Usethebalanceequationprescriptiontoobtainanexpressionforthe“generation

rate”associatedwith Wxx .Solution:

Gφ = −qE x

1Ω

∂φ∂px

fp∑

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

Theterminsidethesumis

∂φ∂px

=∂ 12m*υx

2⎛⎝⎜

⎞⎠⎟

∂px= 12m*

∂ px2( )

∂px= υx

sowefind

Gφ = −qE i

1Ω

υx f!p∑

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪= JnxE x

ECE-656 Fall20178

Exam3Solutions:ECE656Fall20173c) Writedownanexpressionforthe“recombinationrate”associatedwith Wxx .Solution:

Rφ ≡nφ − nφ

0

τφ

=Wxx −Wxx

0

τWxx Wxx

0 = n0

kBTL

2

3d) Writedownthetotalbalanceequationfor Wxx ,andexplainwhatwouldneedtobe

doneinordertosolvethisequationincombinationwiththefirstbalanceequation(thecontinuityequation)andthesecondbalanceequation(themomentumbalanceequation).

Solution:

∂Wxx

∂t= − d

dxFWxx( ) + JnxE x −

Wxx −Wxx0

τWxx

Tosolvethisequation,wewouldneedtoterminatethehierarchyanddevelopanapproximationfor

FWxx

intermsofthequantitiesinthepreviousbalanceequations.

Wewouldalsoneedtodevelopanexpressionfortherelaxationtime,

τWxx

.

ECE-656 Fall20179

Exam3Solutions:ECE656Fall20174) ThisproblemconcernstheballisticSchottkybarrierdiodeshowninthefigure

below.Notethatitisforwardbiased.Assumethatthesemiconductorissiliconwithaneffectivedensityofstatesof NC = 3.23×1019 cm-3 andthatthetemperatureis300K.Thequantity, Fn ,istheelectronquasi-Fermilevel(elecrochemicalpotential)inthesemiconductorand EFM istheFermilevelinthemetal.Notshownontheleftisanidealcontactinequilibrium.

Answerthefollowingquestions.Numericalanswersarerequired.

4a) Atthemetal-semiconductorjunction,whatisthedensityofballisticelectronsinthe

semiconductorwithpositivevelocities, n+ x = 0−( )?

Solution:Inabulk,diffusivesemiconductor,

n = NCe Fn−EC( ) kBT At x = 0− onlythepositivevelocitystatesarefilledbythecontactattheleft.Halfofthestateshavepositivevelocities,so

n+ x = 0−( ) = NC

2e Fn−EC( ) kBT = 3.23×1019

2e−0.1/0.026 = 3.45×1017 cm-3

n+ x = 0−( ) = 3.45×1017 cm-3

Thenon-degenerateassumptionwearemakingisclearlyOK.Exam3Solutions:ECE656Fall2017

ECE-656 Fall201710

4b) Atthemetal-semiconductorjunction,whatisthedensityofballisticelectronsinthe

semiconductorwithnegativevelocities, n− x = 0−( ) ?

Solution:Inabulk,diffusivesemiconductor,

n = NCe Fn−EC( ) kBT

At x = 0− onlythenegativevelocitystatesarefilledbythecontactattheright.Halfofthestateshavenegativevelocities,so

n− x = 0−( ) = NC

2e EFM −EC( ) kBT = 3.23×1019

2e−0.2/0.026 = 7.37 ×1015 cm-3

n− x = 0−( ) = 7.37 ×1015 cm-3

Thenon-degenerateassumptionwearemakingisclearlyOK.

4c) WhatisthecurrentdensityinAmperespercm2?Youwillneedaneffectivemassfor

thisquestion;youmayassumeasimpleparabolicbandwithaneffectivemassof

m* = mC

* = 0.26m0 .Solution:

Jn = qυT n+ 0−( )− n− 0−( )⎡

⎣⎤⎦

Fromtheformulasheet:

υx

+ =υT = 2kBT π m* ηF << 0( )

υT =2kBTπ m* =

2 1.38×10−23( )300

3.14 0.26( ) 9.11×10−31( ) = 1.06×105 m/s

Jn = qυT n+ 0−( )− n− 0−( )⎡

⎣⎤⎦ = 1.6×1019 1.06×107( ) 3.45×1017 − 7.37 ×1015⎡⎣ ⎤⎦ A/ cm2

Jn = 5.73×105 A/ cm2

ECE-656 Fall201711

ECE-656KeyEquations(Weeks1-15)

Physicalconstants:

= 1.055 ×10−34 J-s[ ] m0 = 9.109 ×10

−31 kg[ ] kB = 1.380 ×10

−23 J/K[ ] q = 1.602 ×10−19 C[ ] ε0 = 8.854 ×10−14 F/cm[ ]

--------------------------------------------------------------------------------------------------------------------Densityofstatesink-space:1D:

Nk =2 × L 2π( ) = L π 2D: Nk =2 × A 4π 2( ) = A 2π 2 3D:

Nk =2 × Ω 8π 2( ) = Ω 4π 3 Densityofstatesinenergy(parabolicbands,perlength,area,orvolume):

D1D E( ) = gv

π2m*

E − ε1( ) D2D E( ) = gV

m*

π2 D3D E( ) = gv

m* 2m* E − EC( )π 23

--------------------------------------------------------------------------------------------------------------------FermifunctionandFermi-DiracIntegrals/Bose-Einsteinfunction

f0 E( ) = 1

1+ e E−EF( ) kBT N0 =

1e!ω0 kBT −1

F j ηF( ) = 1

Γ( j +1)η jdη

1+ eη−ηF0

∞

∫ F j ηF( )→ eηF ηF << 0

dF j

dηF

=F j−1

Γ(n) = n −1( )! (naninteger) Γ(1 / 2) = π Γ( p +1) = pΓ( p) --------------------------------------------------------------------------------------------------------------------Scattering:

S !p, ′!p( ) = 2π

"H !′p , !p

2δ ′E − E − ΔE( )

H ′p , p =

1Ω

e− i ′p r

−∞

+∞

∫ US (r )ei pr dr

1τ p( ) == S p, ′p( )

′p ,↑∑

1τ mp( ) = S p, ′p( )Δpz

pz0′p ,↑∑

1τ Ep( ) = S p, ′p( )ΔE

E0′p ,↑∑

ADP: Kq

2= q2DA

2 ODP: Kq

2= D0

2 PZ: Kq

2= eePZ κ Sε0( )2 POP: Kβ

2= ρe2ω 0

2

q2κ 0ε0κ 0

κ∞

−1⎛⎝⎜

⎞⎠⎟

S !p, ′!p( ) = π

Ωρ ωKq

2Nω + 1

2∓

12

⎛⎝⎜

⎞⎠⎟δ ′!p − !p ∓ #!q( )δ ′E − E ∓ #ω( )

δ ′!p − !p ∓ #!q( )δ ′E − E ∓ #ωβ( )→ 1

#υqδ ±cosθ + #q

2 p∓ω q

υq⎛

⎝⎜

⎞

⎠⎟

1τ= 1τ abs

+ 1τ ems

= 2π

DA2kBTcl

⎛

⎝⎜⎞

⎠⎟D3D E( )

2(ADP)

LD =

κ Sε0kBTq2n0

1τ= 1τ m

= 2πDO

2

2ρω0

⎛

⎝⎜⎞

⎠⎟N0 +

12

12

⎛⎝⎜

⎞⎠⎟

D3D E ± ω0( )2

N0 =

1eωo kB T −1

(ODP)

ECE-656 Fall201712

BoltzmannTransportEquation:

∂ f∂t

+!υ i∇r f +

!Fe i∇ p f = df

dt coll

!Fe = −q

!E − q

!υ ×!B

dfdt coll

= −f − fS( )τ m

= −δ fτ m

(RTA)

WithasmallB-field,theresulting2Dcurrentequationis !Jn =σ S

!E -σ SµH

!E ×

!B( )

ω cτ m <<1( )

µH = µnrH rH ≡ τ m

2 τ m

2

• ≡ •( )E E

IsothermalNear-EquilibriumTransport:SummaryofLandauerApproach

I = 2q

hT E( )∫ M E( ) f1 − f2( )dE àsmallbias,isothermal: f1 E( )− f2 E( ) = − ∂ f0

∂E⎛⎝⎜

⎞⎠⎟ qV( )

Linearresponse(alsocalledlowbias,near-equilibrium):

I = GV G = 2q2

hT E( )∫ M E( ) −

∂ f0

∂E⎛⎝⎜

⎞⎠⎟

dE G = 2q2

hT E( ) M E( )

T E( ) ≡T E( )∫ M E( ) −∂ f0 ∂E( )dE

M E( ) −∂ f0 ∂E( )dE∫⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

M E( ) = M E( ) −∂ f0 ∂E( )dE∫

Rball =

1M EF( )

h2q2 =

12.8kΩM

Modes/channels(general):

M E( ) ≡ h4

υx+ E( ) D1D E( ) υx

+ E( ) =υ E( )

M E( ) =WM 2D ≡W h4

υx+ E( ) D2D E( ) υx

+ E( ) = 2πυ E( )

M E( ) = AM 3D ≡ A h4

υx+ E( ) D3D E( ) υx

+ E( ) = 12υ E( )

Modes(Parabolicbands):

E k( ) = EC + !2k 2 2m*( ) Modes(graphene):

M E( ) = M1D E( ) = gv

M E( ) =W M2D E( ) = gvW

2m* E − EC( )π!

M (E) =W 2 E π!υF

M E( ) = A M3D E( ) = gv A m*

2π!2 E − EC( )

Transmission: T E( ) = λ E( )

λ E( ) + L

ECE-656 Fall201713

Mean-free-pathforbackscattering:

1D: λ E( ) = 2υ E( )τ m E( ) 2D: λ E( ) = π2υ E( )τ m E( ) 3D: λ E( ) = 4

3υ E( )τ m E( )

Diffusioncoefficient:

Dn = υx+ λ 2 Dn =υT λ0 2

Dn E( ) = υx+ E( ) λ E( ) 2

Uni-directionalthermalvelocity:

υx

+ =υT = 2kBT π m* ηF << 0( ) Thermoelectrictransport:Coupledcurrentequations(diffusive): Jx =σE x −σ S dT dx Jx

Q = Tσ SE x −κ 0 dT dx

Coupledcurrentequations(inverted): E x = ρJx + S

dTL

dx

Jx

Q = π Jx −κ e

dTdx

Transportcoefficients:

σ = ′σ E( )∫ dE =

2q2

hM λ

′σ E( ) = 2q2

hλ E( ) M E( )

A−∂ f0

∂E⎛⎝⎜

⎞⎠⎟

M = M E( ) −∂f0

∂E⎛

⎝⎜⎞

⎠⎟∫ dE

S = −

kB

qE − EF

kBT⎛

⎝⎜⎞

⎠⎟′σ E( )∫ dE ′σ E( )∫ dE = −

kB

q⎛

⎝⎜⎞

⎠⎟E − EF

kBT=

kB

−q⎛

⎝⎜⎞

⎠⎟EC − EF( )

kBT+

Δn

kBT

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪ π = TLS (KelvinRelation)

κ 0 = TkB

q⎛

⎝⎜⎞

⎠⎟

2E − EF

kBT⎛

⎝⎜⎞

⎠⎟

2

′σ E( )∫ dE =κ 0 = σTkB

q⎛

⎝⎜⎞

⎠⎟

2E − EF

kBT⎛

⎝⎜⎞

⎠⎟

2⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪ave κ e =κ 0 − πSσ

κ e

σ=

kB

q⎛⎝⎜

⎞⎠⎟

2E − EF

kBT⎛

⎝⎜⎞

⎠⎟

2

−E − EF

kBT⎛

⎝⎜⎞

⎠⎟

2⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪T = LT (Weidemann-Franz“Law”)

2

kB

q⎛⎝⎜

⎞⎠⎟

2

< L < π 2

3kB

q⎛⎝⎜

⎞⎠⎟

2

(parabolicbandswithenergy-independentscattering)

ThermoelectricmaterialFigureofMerit: zT =

S 2σTκ

Latticethermalconductivity: κ L =

π 2kB2TL

3hλ ph !ω( )M ph !ω( )

AWph !ω( )d !ω( )∫

Generalbalanceequation

ECE-656 Fall201714

∂nφ∂t

= −∇ •!Fφ +Gφ − Rφ

Prescription:

nφ (!r ,t) = 1

Ωφ( !p) f !r , !p,t( )

!p∑

!Fφ ≡

1Ω

φ !p( ) !υ f !r , !p,t( )!p∑

Gφ = −q

!E i

1Ω

!∇ pφ f

!p∑

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

Rφ ≡

nφ − nφ0

τφ

Currentequation:

Jnx = nqµnE x + 2µn

d nuxx( )dx

u = 1

2m*υd

2 + 32

kBTe

υSAT ≈

!ω0

m*